QJEEM20Ma0cba: JEE Main 2020 Mathematics - Class 11 JEEM20Ma0cba Solution: If α and β be two roots of the equation x 2 – 64x + 256 = 0.…

Question

Verified Solution • 2025–26 Session moderate

If

α

and

β

be two roots of the equation
x² – 64x + 256 = 0. Then the value of

{(\frac{α^{3}}{β^{5}})}^{1 / 8} + {(\frac{β^{3}}{α^{5}})}^{1 / 8}

is :

Accepted Answer

x² – 64x + 256 = 0

α

+

β

= 64,

α

β

= 256

{(\frac{α^{3}}{β^{5}})}^{1 / 8} + {(\frac{β^{3}}{α^{5}})}^{1 / 8}

=

\frac{α^{\frac{3}{8}}}{β^{\frac{5}{8}}} + \frac{β^{\frac{3}{8}}}{α^{\frac{5}{8}}}

=

\frac{α + β}{{(α β)}^{\frac{5}{8}}}

=

\frac{64}{{(256)}^{\frac{5}{8}}}

= 2

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